Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration


Fundamentals Of Physics
Found in: Page 117

Answers without the blur.

Just sign up for free and you're in.


Short Answer

A 0.150 Kg particle moves along an x axis according to x(t)=-13.00+2.00t+4.00t2-3.00t3, with x in meters and t in seconds. In unit-vector notation, what is the net force acting on the particle at t = 3.40 s?

The net force acting on the particle in the unit-vector notation at t = 3.40s is -7.98 Ni^ .

See the step by step solution

Step by Step Solution

Step 1: The given data

  • Mass of the particle, m = 0.150 kg.
  • Time at which we need to calculate the force, t = 3.40s .
  • Position of the particle of mass 0.150 kg along x axis is x(t)=-13.00+2.00t+4.00t2-3.00t3

Step 2: Understanding the concept of Newton’s law of motion

The net force FNet on a body is a product of mass m and acceleration aof the body and is given by,


The acceleration can be differentiating the displacement vector twice with respect to time.

Using the formula for Newton’s second law, we can find the net force acting on the particle at t = 3.40 s


The net force acting on the body,

FNet=Ma (i)

Here, Fnetis the net force, M is mass of the object, and ais the acceleration.

The acceleration of a body in motion, a=d2xdt2 (ii)

Step 3: Calculation of net force

From equation (ii) and the given position equation,


we get the acceleration of the particle from equation (ii) by substituting the displacement in the equation.

role="math" localid="1657015679628" a=d2-13.00+2.00t+4.00t2-3.00t3dt2 =8.00-18.0t =53.2 m/s2 (t= 3.40 s)

Now using equation (i) and the derived and given values, we get the net force as:

FNet=(0.150 kg)(8.00-18.0t) =(-7.98 N)i^ (t= 3.40 s)

Hence, the net force in unit-vector notation is (-7.98N)i^.

Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.